Why do such different Lagrangians give the same Euler-Lagrange equations?

Question

The Lagrangians $$L_\pm:=\frac{1}{2}\left( \dot{q}_1^2\pm\dot{q}_2^2\right)-\frac{1}{2}m^2\left( q_1^2\pm q_2^2\right)$$ each have Euler-Lagrange equations $\ddot{q}_i=-m^2q_i,\,i\in\left\{ 1,\,2\right\}$. The obvious reason is that these equations are uncoupled, so any linear combination of "one-equation" Lagrangians, including these $\pm$ choices, will give the same pair.

The Lagrangian $L_0:=\dot{q}_1\dot{q}_2-m^2q_1q_2$ has the same ELEs as well, so it doesn't couple the $q_i$ even though it looks like it might. What's the physical interpretation of $L_0$ obtaining the same ELEs as $L_+$? It doesn't seem to be of the form $a_+L_++a_-L_-+b+\dot{f}$.

Thinking in terms of Hamiltonians doesn't make it any clearer. With $L_0$ as our choice of Lagrangian, the Hamiltonian is $H_0:=p_1p_2+m^2q_1q_2$, which bears no obvious equivalence to $$H_\pm:=\frac{1}{2}\left( p_1^2\pm p_2^2\right)+\frac{1}{2}m^2\left( q_1^2\pm q_2^2\right)$$ (The momenta have different definitions in terms of the $\dot{q}_i$ in the two cases, but of course we still get the same equations of motion in all cases.)

So is there a more general principle that explains why $L_\pm$ is equivalent to $L_0$, or $H_\pm$ to $H_0$?

Answer 1

Having thought about this, I think I've found the key point. A Lagrangian of the form $\frac{1}{2}A\dot{q}^2-\frac{1}{2}Bq^2$ with $1$-dimensional $q$ and numbers $A,\,B$ with $A\ne 0$ has equations of motion that depend only on $A^{-1}B$, as scaling the Lagrangian is physically irrelevant. In particular, scaling the Lagrangian can be thought of as a scaling of the coefficients. For multi-dimensional $\mathbf{q}$, we can generalise this result. In a Lagrangian $L_{AB}:=\frac{1}{2}\dot{\mathbf{q}}^TA\dot{\mathbf{q}}-\frac{1}{2}\mathbf{q}^TB\mathbf{q}$ we can assume without loss of generality that $A,\,B$ are symmetric, in which case $\mathbf{p}=A\dot{\mathbf{q}}$ so the equation of motion is $A\ddot{\mathbf{q}}=-B\mathbf{q}$. If $A$ is invertible this simplifies to $\ddot{\mathbf{q}}=-A^{-1}B\mathbf{q}$, which is invariant under a "scaling" of the matrix coefficients of the form $A,\,B\to KA,\,KB$, with $K$ an arbitrary invertible square matrix conformable with $A,\,B$ for which $KA,\,KB$ are symmetric. We can find the same result in the Hamiltonian formalism, viz $H_{AB}:=\frac{1}{2}\mathbf{p}^TA^{-1}\mathbf{p}+\frac{1}{2}\mathbf{q}^TB\mathbf{q}$.